Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 50;
const int oo = 1e9;
const int mod = 1e9 + 7;
vector<int> g[N], all[N][18];
int d[N], p[N][18];
void dfs(int u, int q) {
  p[u][0] = q;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v == p[u][0]) continue;
    d[v] = d[u] + 1;
    dfs(v, u);
  }
}
set<int> lca(int u, int v) {
  set<int> ans;
  if (d[u] < d[v]) swap(u, v);
  int l = d[u] - d[v];
  for (int k = 17; k >= 0; k--) {
    if ((l >> k) & 1) {
      for (int i = 0; i < all[u][k].size(); i++) {
        ans.insert(all[u][k][i]);
      }
      u = p[u][k];
    }
  }
  for (int i = 0; i < all[u][0].size(); i++) {
    ans.insert(all[u][0][i]);
  }
  for (int i = 0; i < all[v][0].size(); i++) {
    ans.insert(all[v][0][i]);
  }
  if (v == u) return ans;
  for (int k = 17; k >= 0; k--) {
    if (p[u][k] != p[v][k]) {
      for (int i = 0; i < all[u][k].size(); i++) {
        ans.insert(all[u][k][i]);
      }
      for (int i = 0; i < all[v][k].size(); i++) {
        ans.insert(all[v][k][i]);
      }
      u = p[u][k];
      v = p[v][k];
    }
  }
  for (int i = 0; i < all[u][0].size(); i++) {
    ans.insert(all[u][0][i]);
  }
  for (int i = 0; i < all[v][0].size(); i++) {
    ans.insert(all[v][0][i]);
  }
  for (int i = 0; i < all[p[v][0]][0].size(); i++) {
    ans.insert(all[p[v][0]][0][i]);
  }
  return ans;
}
int main() {
  memset(p, -1, sizeof(p));
  int n, m, q;
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 1; i <= m; i++) {
    int a;
    scanf("%d", &a);
    if (all[a][0].size() < 10) all[a][0].push_back(i);
  }
  dfs(1, -1);
  for (int i = 1; i < 18; i++) {
    for (int j = 1; j <= n; j++) {
      if (p[j][i - 1] == -1) continue;
      int p2 = p[j][i - 1];
      p[j][i] = p[p2][i - 1];
      vector<int> temp;
      for (int l = 0; l < all[j][i - 1].size(); l++) {
        temp.push_back(all[j][i - 1][l]);
      }
      for (int l = 0; l < all[p2][i - 1].size(); l++) {
        temp.push_back(all[p2][i - 1][l]);
      }
      sort(temp.begin(), temp.end());
      for (int l = 0; l < temp.size(); l++) {
        if (l == 10) break;
        all[j][i].push_back(temp[l]);
      }
    }
  }
  while (q--) {
    int v, u, a;
    scanf("%d%d%d", &v, &u, &a);
    if (d[u] < d[v]) swap(u, v);
    set<int> ans = lca(u, v);
    int q = ans.size();
    q = min(q, a);
    printf("%d ", q);
    int cnt = 0;
    for (set<int>::iterator it = ans.begin(); it != ans.end(); it++) {
      if (cnt++ == q) break;
      printf("%d ", *it);
    }
    puts("");
  }
}